可分离变量微分方程的两种形式：导数形式和微分形式

一、前言

对于可分离变量的微分方程，我们如下这种导数形式的表达式：

$$
\frac{\mathrm{d} y}{\mathrm{d} x} = f(x)g(y)
$$

以及下面这种微分形式的表达式：

$$
f_{1}(x) g_{1}(y) \mathrm{~d} x + f_{2}(x) g_{2}(y) \mathrm{~d} y = 0
$$

事实上，上面这两种表达形式是完全等价的.

接下来，荒原之梦考研数学就通过代数运算，证明这二者的等价关系.

二、正文

2.1 微分形式 $\Rightarrow$ 导数形式

$$
\begin{aligned}
& \textcolor{lightgreen}{ f_{1}(x) g_{1}(y) \mathrm{~d} x + f_{2}(x) g_{2}(y) \mathrm{~d} y = 0 } \\ \\
\leadsto \ & f_{2}(x) g_{2}(y) \mathrm{~d} y = -f_{1}(x) g_{1}(y) \mathrm{~d} x \\ \\
\leadsto \ & \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{-f_{1}(x) g_{1}(y)}{f_{2}(x) g_{2}(y)}, \ \textcolor{gray}{f_{2}(x)g_{2}(y) \neq 0} \\ \\
\leadsto \ & \frac{\mathrm{d} y}{\mathrm{d} x} = \left[ -\frac{f_{1}(x)}{f_{2}(x)} \right] \cdot \left[ \frac{g_{1}(y)}{g_{2}(y)} \right] \\ \\
\leadsto \ & \textcolor{gray}{F(x) = -\frac{f_{1}(x)}{f_{2}(x)}, \ G(y) = \frac{g_{1}(y)}{g_{2}(y)}} \\ \\
\leadsto \ & \textcolor{lightgreen}{ \frac{\mathrm{d} y}{\mathrm{d} x} = F(x)G(y) }
\end{aligned}
$$

对于 $f_{1}(x) g_{1}(y) \mathrm{~d} x$ $+$ $f_{2}(x) g_{2}(y) \mathrm{~d} y$ $=$ $0$，分离变量并求解的方式为：

$$
\begin{aligned}
& \frac{g_{2}(y)}{g_{1}(y)} \mathrm{~d} y = -\frac{f_{1}(x)}{f_{2}(x)} \mathrm{~d} x \\ \\
\leadsto \ & \int \frac{g_{2}(y)}{g_{1}(y)} \mathrm{~d} y = – \int \frac{f_{1}(x)}{f_{2}(x)} \mathrm{~d} x
\end{aligned}
$$

2.2 导数形式 $\Rightarrow$ 微分形式

$$
\begin{aligned}
& \textcolor{lightgreen}{ \frac{\mathrm{d} y}{\mathrm{d} x} = f(x)g(y) } \\ \\
\leadsto \ & \mathrm{~d} y = f(x)g(y) \mathrm{~d} x \\ \\
\leadsto \ & f(x)g(y) \mathrm{~d} x – 1 \cdot \mathrm{~d} y = 0 \\ \\
\leadsto \ & \textcolor{gray}{f_{1}(x) = f(x), \ g_{1}(y) = g(y), \ f_{2}(x) = 1, \ g_{2}(y) = -1} \\ \\
\leadsto \ & \textcolor{lightgreen}{ f_{1}(x) g_{1}(y) \mathrm{~d} x + f_{2}(x) g_{2}(y) \mathrm{~d} y = 0 }
\end{aligned}
$$

对于 $\frac{\mathrm{d} y}{\mathrm{d} x}$ $=$ $f(x)g(y)$, 分离变量并求解的方式为：

$$
\begin{aligned}
& \frac{1}{g(y)} \mathrm{~d} y = f(x) \mathrm{~d} x \\ \\
\leadsto \ & \int \frac{1}{g(y)} \mathrm{~d} y = \int f(x) \mathrm{~d} x
\end{aligned}
$$

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